Problem 4 - Elementary Algebra

Question

For Problems \(1-12\), solve each of the equations. These equations are the types you will be using in Problems 13-40. $$ \ell+\frac{2}{3} \ell+1=41 $$

Step-by-Step Solution

Answer

The solution is \( \ell = 24 \).

Step 1: Combine like terms

The given equation is \( \ell + \frac{2}{3} \ell + 1 = 41 \). First, we need to simplify the expression on the left side of the equation by combining like terms. The like terms are \( \ell \) and \( \frac{2}{3} \ell \). We can combine them as follows: \( 1\ell + \frac{2}{3}\ell = \frac{5}{3}\ell \). So, the equation becomes \( \frac{5}{3}\ell + 1 = 41 \).

Step 2: Isolate the variable term

Our goal is to solve for \( \ell \), so we need to isolate the term containing the variable on one side of the equation. We do this by subtracting 1 from both sides of the equation: \( \frac{5}{3}\ell + 1 - 1 = 41 - 1 \), which simplifies to \( \frac{5}{3}\ell = 40 \).

Step 3: Solve for \( \ell \)

Now, we need to solve for \( \ell \) by isolating it completely. Since \( \frac{5}{3}\ell = 40 \), we can multiply both sides by the reciprocal of \( \frac{5}{3} \), which is \( \frac{3}{5} \), to solve for \( \ell \): \( \ell = 40 \times \frac{3}{5} \). Calculate the right side: \( \ell = \frac{120}{5} = 24 \). Therefore, \( \ell = 24 \).

Key Concepts

Combining Like Terms

In algebra, simplifying expressions is a key step to solving equations, and it often begins with combining like terms. Like terms are terms that have the same variable raised to the same power. In the given problem, the equation is \( \ell + \frac{2}{3} \ell + 1 = 41 \). To combine like terms in this context:

• The like terms are \( \ell \) and \( \frac{2}{3} \ell \) because both contain the variable \( \ell \).
• We add their coefficients: \( 1 \times \ell + \frac{2}{3} \times \ell = \frac{5}{3} \ell \).

This simplifies the left side of the equation to \( \frac{5}{3} \ell + 1 \), reorganizing the expression to make further steps in solving the equation more straightforward.

Isolating Variables

The next step in solving an equation involves isolating the variable, which means getting the variable by itself on one side of the equation. This process moves us closer to finding the value of the variable. For our equation \( \frac{5}{3} \ell + 1 = 41 \):

• We need to eliminate the constant \( 1 \) from the left side by subtracting it from both sides: \( \frac{5}{3} \ell + 1 - 1 = 41 - 1 \).
• The equation then simplifies to \( \frac{5}{3} \ell = 40 \).

This action has effectively isolated the term with \( \ell \), setting us up to find its exact value in the next step.

Solving Linear Equations

Now that we have the equation \( \frac{5}{3} \ell = 40 \), solving for \( \ell \) involves finding its exact value. This involves the mathematical operation of division or multiplication:

• To solve for \( \ell \), we can multiply each side by the reciprocal of \( \frac{5}{3} \), which is \( \frac{3}{5} \).
• This step effectively cancels out the fraction and isolates \( \ell \) on the left side: \( \frac{3}{5} \times \frac{5}{3} \ell = 40 \times \frac{3}{5} \).
• This results in \( \ell = 24 \).

Using reciprocal multiplication ensures that the coefficient of \( \ell \) becomes 1, effectively solving this linear equation and demonstrating that the solution to the original problem is \( \ell = 24 \).